18 - Irreducible Tensor Operators and the Wigner-Eckart Theorem.pdf

8/10/2019 18 - Irreducible Tensor Operators and the Wigner-Eckart Theorem.pdf

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Physics 221A

Fall 2011

Notes 18

Irreducible Tensor Operators and the

Wigner-Eckart Theorem

1. Introduction

The Wigner-Eckart theorem concerns matrix elements of a type that is of frequent occurrencein all areas of quantum physics, especially in perturbation theory and in the theory of the emis-

sion and absorption of radiation. This theorem allows one to determine very quickly the selection

rules for the matrix element that follow from rotational invariance. In addition, if matrix elements

must be calculated, the Wigner-Eckart theorem frequently offers a way of significantly reducing the

computational effort. We will make quite a few applications of the Wigner-Eckart theorem in this

course, including several in the second semester.

The Wigner-Eckart theorem is based on an analysis of how operators transform under rotations.

It turns out that operators of a certain type, the irreducible tensor operators, are associated with

angular momentum quantum numbers and have transformation properties similar to those of kets

with the same quantum numbers. An exploitation of these properties leads to the Wigner-Eckarttheorem.

2. Definition of a Rotated Operator

We consider a quantum mechanical system with a ket space upon which rotation operators U(R),

forming a representation of the classical rotation group SO(3), are defined. The representation may

be double-valued, as it will be in the case of systems of half-integer angular momentum. In these

notes we consider only proper rotations R; improper rotations will be taken up later. The operators

U(R) map kets into new or rotated kets,

|

= U(R)

|

, (1)

where| is the rotated ket. We will also write this as| U(R)|. (2)

In the case of half-integer angular momenta, the mapping above is only determined to within a sign

by the classical rotation R.

Now ifA is an operator, we define the rotated operatorA by requiring that the expectation

value of the original operator with respect to the initial state be equal to the expectation value of

the rotated operator with respect to the rotated state, that is,

|A|=|A|, (3)


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2 Notes 18: Irreducible Tensor Operators

which is to hold for all initial states|. But this implies|U(R) A U(R)|=|A|, (4)

or, since| is arbitrary (see Prob. 1.6(b)),U(R) A U(R) = A. (5)

Solving for A, this becomes

A =U(R) A U(R), (6)

which is our definition of the rotated operator. We will also write this in the form,

A

U(R) A U(R). (7)

Notice that in the case of half-integer angular momenta the rotated operator is specified by the

SO(3) rotation matrix R alone, since the sign ofU(R) cancels and the answer does not depend on

which of the two rotation operators is used on the right hand side. This is unlike the case of rotating

kets, where the sign does matter.

3. Scalar Operators

Now we classify operators by how they transform under rotations. First we define a scalar

operatorKto be an operator that is invariant under rotations, that is, that satisfies

U(R

) K U(R

)

=K, (8)

for all operators U(R). This terminology is obvious. Notice that it is equivalent to the statement

that a scalar operator commutes with all rotations,

[U(R), K] = 0. (9)

If an operator commutes with all rotations, then it commutes in particular with infinitesimal rota-

tions, and hence with the generators J. See Eq. (12.13). Conversely, if an operator commutes with

J (all three components), then it commutes with any function ofJ, such as the rotation operators.

Thus another equivalent definition of a scalar operator is one that satisfies

[J, K] = 0.

(10)

The most important example of a scalar operator is the Hamiltonian for an isolated system, not

interacting with any external fields. The consequences of this for the eigenvalues and eigenstates

of the Hamiltonian are discussed in Sec. 15 below, where it is shown that the energy eigenspaces

consist of one or more irreducible subspaces under rotations. In fact, apart from exceptional cases

like the electrostatic model of hydrogen, each energy eigenspace consists of precisely one irreducible

subspace. Hence the eigenspaces are characterized by an angular momentum quantum number j

and the energy eigenvalue is (2j+ 1)-fold degenerate.


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Notes 18: Irreducible Tensor Operators 3

4. Vector Operators

In ordinary vector analysis in three-dimensional Euclidean space, a vector is defined as a collec-

tion of three numbers that have certain transformation properties under rotations. It is not sufficient

just to have a collection of three numbers; they must in addition transform properly. Similarly, in

quantum mechanics, we define a vector operatoras a vector ofoperators (that is, a set of three

operators) with certain transformation properties under rotations.

Our requirement shall be that the expectation value of a vector operator, which is a vector of

ordinary orc-numbers, should transform as a vector in ordinary vector analysis. This means that if

|

is a state and

|

is the rotated state as in Eq. (1), then

|V|= R|V|, (11)

where V is the vector of operators that qualify as a genuine vector operator. In case the notation

in Eq. (11) is not clear, we write the same equation out in components,

|Vi|=j

Rij|Vj |. (12)

Equation (11) or (12) is to hold for all|, so by Eq. (1) they imply (after swapping R and R1)

U(R) V U(R) = R1V, (13)

or, in components,U(R) Vi U(R)

=j

VjRji.

(14)

We will take Eq. (13) or (14) as the definition of vector operator.

In the case of a scalar operator, we had one definition (8) involving its properties under conjuga-

tion by rotations, and another (10) involving its commutation relations with the angular momentum

J. The latter is in effect a version of the former, when the rotation is infinitesimal. Similarly, for

vector operators there is a definition equivalent to Eq. (13) or (14) that involves commutation rela-

tions with J. To derive it we let U and R in Eq. (13) have axis-angle form with an angle 1, sothat

U(R) = 1 ih

n J, (15)and

R= I + n J. (16)Then the definition (13) becomes

1 i

hn J

V

1 + i

hn J

= (I n J)V, (17)

or

[n J, V] =ihnV. (18)


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Taking the j -th component of this, we have

ni[Ji, Vj ] =ih jikniVk , (19)

or, sincen is an arbitrary unit vector,

[Ji, Vj ] =ih ijkVk.(20)

Any vector operator satisfies this commutation relation with the angular momentum of the system.

The converse is also true; if Eq. (20) is satisfied, then V is a vector operator. This follows since

Eq. (20) implies Eq. (18) which implies Eq. (17), that is, it implies that the definition (13) is satisfied

for infinitesimal rotations. But it is easy to show that if Eq. (13) is true for two rotations R1 and

R2, then it is true for the product R1R2. Therefore, since finite rotations can be built up as the

product of a large number of infinitesimal rotations (that is, as a limit), Eq. (20) implies Eq. (13)

for all rotations. Equations (13) and (20) are equivalent ways of defining a vector operator.

We have now defined scalar and vector operators. Combining them, we can prove various

theorems. For example, if V and W are vector operators, then VW is a scalar operator, andVWis a vector operator. This is of course just as in vector algebra, except that we must rememberthat operators do not commute, in general. For example, it is not generally true that V W= W V,or that V

W=

W

V.

If we wish to show that an operator is a scalar, we can compute its commutation relations with

the angular momentum, as in Eq. (10). However, it may be easier to consider what happens when

the operator is conjugated by rotations. For example, the central force Hamiltonian (16.1) is a scalar

because it is a function of the dot products p p= p2 andx x= r2. See Sec. 16.2.

5. Tensor Operators

Finally we define a tensor operatoras a tensor ofoperators with certain transformation prop-

erties that we will illustrate in the case of a rank-2 tensor. In this case we have a set of 9 operators

Tij , where i, j = 1, 2, 3, which can be thought of as a 3 3 matrix of operators. These are required

to transform under rotations according to

U(R) TijU(R) =

k

Tk RkiRj, (21)

which is a generalization of Eq. (14) for vector operators. As with scalar and vector operators, a

definition equivalent to Eq. (21) may be given that involves the commutation relations ofTij with

the components of angular momentum.

As an example of a tensor operator, let V and W be vector operators, and write

Tij =ViWj . (22)


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Then Tij is a tensor operator (it is the tensor product ofV with W). This is just an example; in

general, a tensor operator cannot be written as the product of two vector operators as in Eq. (22).

Another example of a tensor operator is the quadrupole moment operator. In a system with a

collection of particles with positions xand charges q, whereindexes the particles, the quadrupole

moment operator is

Qij =

q(3xi xj r2 ij). (23)

It is especially important in nuclear physics, in which the particles are the protons in a nucleus

with charge q= e. Notice that the first term under the sum is an operator of the form (22), with

V= W = x.Tensor operators of other ranks (besides 2) are possible; a scalar is considered a tensor operator

of rank 0, and a vector is considered a tensor of rank 1. In the case of tensors of arbitrary rank, the

transformation law involves one copy of the matrix R1 = Rt for each index of the tensor.

6. Examples of Vector Operators

Consider a system consisting of a single spinless particle moving in three-dimensional space, for

which the wave functions are (x) and the angular momentum is L= xp. To see whether x is avector operator (we expect it is), we compute the commutation relations with L, finding,

[Li, xj ] =ih ijkxk. (24)

According to Eq. (20), this confirms our expectation. Similarly, we find

[Li, pj] =ih ijkpk, (25)

so that p is also a vector operator. Then xp (see Sec. 4) must also be a vector operator, that is,we must have

[Li, Lj] =ih ijkLk. (26)

This last equation is of course just the angular momentum commutation relations, but here with a

new interpretation. More generally, by comparing the adjoint formula (13.80) with the commutation

relations (20), we see that J is a vector operator on any Hilbert space upon which the angularmomentum is defined.

7. The Spherical Basis

The spherical basis is a basis of unit vectors in ordinary three-dimensional space that is alter-

native to the usual Cartesian basis. Initially we just present the definition of the spherical basis

without motivation, and then we show how it can lead to some dramatic simplifications in certain

problems. Then we explain its deeper significance. The spherical basis will play an important role

in the development of later topics concerning operators and their transformation properties.


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We denote the usual Cartesian basis byci, i = 1, 2, 3, so that

c1 = x, c2 = y, c3 =z. (27)

We have previously denoted this basis by ei, but in these Notes we reserve the symbol e for the

spherical basis.

The spherical basis is defined by

e1=x + iy2

,

e0= z,

e1= x i

y2 . (28)

This is a complex basis, so vectors with real components with respect to the Cartesian basis have

complex components with respect to the spherical basis. We denote the spherical basis vectors

collectively byeq, q= 1, 0, 1.The spherical basis vectors have the following properties. First, they are orthonormal, in the

sense that

eq eq =qq . (29)Next, an arbitrary vector X can be expanded as a linear combination of the vectors eq ,

X=

q

eq Xq, (30)

where the expansion coefficients are

Xq = eq X. (31)These equations are equivalent to a resolution of the identity in 3-dimensional space,

I=q

eqeq, (32)

in which the juxtaposition of the two vectors represents a tensor product or dyad notation.

You may wonder why we expand X as a linear combination of eq , instead of eq. The lattertype of expansion is possible too, that is, any vector Y can be written

Y= q

eqYq, (33)

where

Yq =eq Y. (34)

These relations correspond to a different resolution of the identity,

I=q

eqeq . (35)

The two types of expansion give the contravariant and covariant components of a vector with respect

to the spherical basis; in this course, however, we will only need the expansion indicated by Eq. (30).


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Now the matrix elements (37) become a product of a radial integral times an angular integral,

nm|xq|nm= 0

r2 dr Rn(r)rRn(r)

4

3

d Ym(, )Y1q(, )Ym(, ).

(41)

We see that all the dependence on the three magnetic quantum numbers (m, q , m) is contained in

the angular part of the integral. Moreover, the angular integral can be evaluated by the three-Ym

formula, Eq. (17.43), whereupon it becomes proportional to the Clebsch-Gordan coefficient,

m|1mq. (42)

The radial integral is independent of the three magnetic quantum numbers (m, q , m), and the trick

we have just used does not help us to evaluate it. But it is only one integral, and after it has

been done, all the other integrals can be evaluated just by computing or looking up Clebsch-Gordan

coefficients.

The selection rule m = q+m in the Clebsch-Gordan coefficient (42) means that many of the

integrals vanish, so we have exaggerated the total number of integrals that need to be done. But had

we worked with the Cartesian components xi ofx, this selection rule might not have been obvious.

In any case, even with the selection rule, there may still be many nonzero integrals to be done (nine,

in the case 3d 2p).The example we have just given of simplifying the calculation of matrix elements for a dipole

transition is really an application of the Wigner-Eckart theorem, which we take up later in these

notes.

9. Significance of the Spherical Basis

To understand the deeper significance of the spherical basis we examine Table 1. The first

row of this table summarizes the principal results obtained in Notes 13, in which we worked out

the matrix representations of angular momentum and rotation operators. To review those results,

we start with a ket space upon which proper rotations act by means of unitary operators U(R), as

indicated in the second column of the table. We refer only to proper rotations R SO(3), andwe note that the representation may be double-valued. The rotation operators have generators,

defined by Eq. (12.13), that is, that equation can be taken as the definition ofJ when the rotation

operators U(R) are given. (Equation (12.11) is equivalent.) The components ofJ satisfy the usual

commutation relations (12.24) since the operators U(R) form a representation of the rotation group.

Next, since J2 and Jz commute, we construct their simultaneous eigenbasis, with an extra index

to resolve degeneracies. Also, we require states with differentm but the sameandj to be related

by raising and lowering operators. This creates the standard angular momentum basis (SAMB),

indicated in the fourth column. In the last column, we show how the vectors of the standard angular


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momentum basis transform under rotations. A basis vector|j m, when rotated, produces a linearcombination of other basis vectors for the same values of and j but different values ofm. This

implies that the space spanned by|j mfor fixedand j , but form =j, . . . , +j is invariant underrotations. This space has dimensionality 2j+ 1. It is, in fact, an irreducible invariant space (more

on irreducible subspaces below). One of the results of the analysis of Notes 13 is that the matrices

Djmm(U) are universal matrices, dependent only on the angular momentum commutation relations

and otherwise independent of the nature of the system.

Space Action Ang Mom SAMB Action on SAMB

Kets | U| J |j m U|j m=m

|j mDjmm

3D Space x Rx iJ eq Req =q

eqD1qq

Operators AU AU . . . T kq U TkqU =q

TkqDkqq

Table 1. The rows of the table indicate different vector spaces upon which rotations act by means of unitary operators.The first row refers to a ket space (a Hilbert space of a quantum mechanical system), the second to ordinary three-dimensional space (physical space), and the third to the space of operators. The operators in the third row are theusual linear operators of quantum mechanics that act on the ket space, for example, the Hamiltonian. The first columnidentifies the vector space. The second column shows how rotations RSO(3) act on the given space. The third columnshows the generators of the rotations, that is, the 3-vector of Hermitian operators that specify infinitesimal rotations.

The fourth column shows the standard angular momentum basis (SAMB), and the last column, the transformation lawof vectors of the standard angular momentum basis under rotations.

At the beginning of Notes 13 we remarked that the analysis of those Notes applies to other

spaces besides ket spaces. All that is required is that we have a vector space upon which rotations

act by means of unitary operators. For other vectors spaces the notation may change (we will not

call the vectors kets, for example), but otherwise everything else goes through.

The second row of Table 1 summarizes the case in which the vector space is ordinary three-

dimensional (physical) space. Rotations act on this space by means of the matrices R, which, being

orthogonal, are also unitary (an orthogonal matrix is a special case of a unitary matrix). The action

consists of just rotating vectors in the usual sense, as indicated in the second column.The generators of rotations in this case must be a vector J of Hermitian operators, that is,

Hermitian matrices, that satisfy

U(n, ) = 1 ih

n J, (43)when is small. Here U really means the same thing as R, since we are speaking of the action

on three-dimensional space, and 1 means the same as the identity matrix I. We will modify this

definition ofJ slightly by writing J = J/h, thereby absorbing the h into the definition of J and

making J dimensionless. This is appropriate when dealing with ordinary physical space, since it

has no necessary relation to quantum mechanics. (The spherical basis is also useful in classical


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mechanics, for example.) Then we will drop the prime, and just remember that in the case of this

space, we will use dimensionless generators. Then we have

U(n, ) = 1 in J. (44)But this is equal to

R(n, ) = I + n J, (45)as in Eq. (11.32), where the vector of matrices J is defined by Eq. (11.22). These imply

J= iJ, (46)

as indicated in the third column of Table 1. Writing out the matrices Ji explicitly, we have

J1 =

0 0 00 0 i

0 i 0

, J2=

0 0 i0 0 0

i 0 0

, J3 =

0 i 0i 0 0

0 0 0

. (47)

These matrices are indeed Hermitian, and they satisfy the dimensionless commutation relations,

[Ji, Jj ] =iijkJk, (48)

as follows from Eqs. (11.34) and (46).

We can now construct the standard angular momentum basis on three-dimensional space. In

addition to Eq. (47), we need the matrices for J2 andJ. These are

J2 = 2 0 00 2 0

0 0 2

(49)

and

J =

0 0 10 0 i

1 i 0

. (50)

We see that J2 = 2I, which means that every vector in ordinary space is an eigenvector ofJ2 with

eigenvalue j (j+ 1) = 2, that is, with j = 1. An irreducible subspace with j = 1 in any vector space

must be 3-dimensional, but in this case the entire space is 3-dimensional, so the entire space consists

of a single irreducible subspace under rotations withj = 1.

The fact that physical space carries the angular momentum value j = 1 is closely related to thefact that vector operators are irreducible tensor operators of order 1, as explained below. It is also

connected with the fact that the photon, which is represented classically by the vector field A(x)

(the vector potential), is a spin-1 particle.

Since every vector in three-dimensional space is an eigenvector J2, the standard basis consists

of the eigenvectors of J3, related by raising and lowering operators (this determines the phase

conventions of the vectors, relative to that of the stretched vector). But we can easily check that

the spherical unit vectors (28) are the eigenvectors ofJ3, that is,

J3eq = qeq, q= 0, 1. (51)


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Furthermore, it is easy to check that these vectors are related by raising and lowering operators,

that is,

Jeq =

(1 q)(1 q+ 1) eq1, (52)whereJis given by Eq. (50). Only the overall phase of the spherical basis vectors is not determined

by these relations. The overall phase chosen in the definitions (28) has the nice feature thate0 =z.

Since the spherical basis is a standard angular momentum basis, its vectors must transform

under rotations according to Eq. (13.79), apart from notation. Written in the notation appropriate

for three-dimensional space, that transformation law becomes

Req =q

eqD1

qq(R). (53)

We need not prove this as an independent result; it is just a special case of Eq. (13.79). This

transformation law is also shown in the final column of Table 1, in order to emphasize its similarity

to related transformation laws on other spaces.

Equation (53) has an interesting consequence, obtained by dotting both sides with eq . We usea round bracket notation for the dot product on the left hand side, and we use the orthogonality

relation (29) on the right hand side, which picks out one term from the sum. We find

eq , Req

=D1qq(R), (54)

which shows that D1qq is just the matrix representing the rotation operator on three-dimensional

space with respect to the spherical basis. The usual rotation matrix contains the matrix elements

with respect to the Cartesian basis, that is,

ci, Rcj

=Rij . (55)

See Eq. (11.7). For a given rotation, matrices R and D1 are similar (they differ only by a change of

basis).

10. Reducible and Irreducible Spaces of Operators

In the third row of Table 1 we consider the vector space of operators. The operators in question

are the operators that act on the ket space of our quantum mechanical system, that is, they are

the usual operators of quantum mechanics, for example, the Hamiltonian. Linear operators can

be added and multiplied by scalars, so they form a vector space in the mathematical sense, but of

course they also act on vectors (that is, kets). So the word vector is used in two different senses

here. Rotations act on operators according to our definition (6), also shown in the second column of

the table. Thus we have another example of a vector space upon which rotation operators act, and

we can expect that the entire construction of Notes 13 will go through again, apart from notation.

Rather than filling in the rest of the table, however, let us return to the definition of a vector

operator, Eq. (14), and interpret it in a different light. That definition concerns the three components


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V1, V2 and V3 of a vector operator, each of which is an operator itself, and it says that if we rotate

any one of these operators, we obtain a linear combination of the same three operators. Thus, any

linear combination of these three operators is mapped into another such linear combination by any

rotation, or, equivalently, the space of operators spanned by these three operators is invariant under

rotations. Thus we view the three components ofV as a set of basis operators spanning this space,

which is a 3-dimensional subspace of the space of all operators. (We assume V= 0.) A generalelement of this subspace of operators is an arbitrary linear combination of the three basis operators,

that is, it has the form

a1V1+ a2V2+ a3V3 = a V, (56)a dot product of a vector of numbers a and a vector of operators V.

If a subspace of a vector space is invariant under rotations, then we may ask whether it contains

any smaller invariant subspaces. If not, we say it is irreducible. If so, it can be decomposed into

smaller invariant subspaces, and we say it is reducible. The invariant subspaces of a reducible space

may themselves be reducible or irreducible; if reducible, we decompose them further. We continue

until we have only irreducible subspaces. Thus, every invariant subspace can be decomposed into

irreducible subspaces, which in effect form the building blocks of any invariant subspace.

In the case of a ket space, the subspaces spanned by|j mfor fixedand j but m=j, . . . , +jis, in fact, an irreducible subspace. The proof of this will not be important to us, but it is not

hard. What about the three-dimensional space of operators spanned by the components of a vector

operator? It turns out that it, too, is irreducible.A simpler example of an irreducible subspace of operators is afforded by any scalar operatorK.

IfK= 0, Kcan be thought of as a basis operator in a one-dimensional space of operators, in whichthe general element isaK, whereais a number (that is, the space contains all multiples ofK). Since

Kis invariant under rotations (see Eq. (8)), this space is invariant. It is also irreducible, because a

one-dimensional space contains no smaller subspace, so if invariant it is automatically irreducible.

We see that both scalar and vector operators are associated with irreducible subspaces of op-

erators. What about second rank tensor operators Tij? Such an operator is really a tensor of

operators, that is, 9 operators that we can arrange in a 3 3 matrix. Assuming these operators arelinearly independent, they span a 9-dimensional subspace of operators that is invariant under rota-

tions, since according to Eq. (21) when we rotate any of these operators we get a linear combinationof the same operators. This space, however, is reducible.

To see this, let us take the example (22) of a tensor operator, that is, Tij = ViWj where V

and W are vector operators. This is not the most general form of a tensor operator, but it will

illustrate the points we wish to make. A particular operator in the space of operators spanned by

the componentsTij is the trace ofTij, that is,

tr T=T11+ T22+ T33 = V W. (57)Being a dot product of two vectors, this is a scalar operator, and is invariant under rotations.

Therefore by itself it spans a 1-dimensional, irreducible subspace of the 9-dimensional space of


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operators spanned by the components ofTij . The remaining (orthogonal) 8-dimensional subspace

can be reduced further, for it possesses a 3-dimensional invariant subspace spanned by the operators,

X3 = T12 T21 = V1W2 V2W1,X1 = T23 T32 = V2W3 V3W2,X2 = T31 T13 = V3W1 V1W3, (58)

or, in other words,

X= VW. (59)

The components ofX form a vector operator, so by themselves they span an irreducible invariantsubspace under rotations. As we see, the components ofX contain the antisymmetric part of the

original tensor Tij.

The remaining 5-dimensional subspace is irreducible. It is spanned by operators containing the

symmetric part of the tensor Tij, with the trace removed (or, as we say, the symmetric, traceless

part ofTij). The following five operators form a basis in this subspace:

S1= T12+ T21,

S2= T23+ T32,

S3= T31+ T13,

S4 = T11 T22,S5 = T11+ T22 2T33. (60)

The original tensor Tij breaks up in three irreducible subspaces, a 1-dimensional scalar (the

trace), a 3-dimensional vector (the antisymmetric part), and the 5-dimensional symmetric, traceless

part. Notice that these dimensionalities are in accordance with the Clebsch-Gordan decomposition,

1 1 = 0 1 2, (61)

which corresponds to the count of dimensionalities,

3 3 = 1 + 3 + 5 = 9. (62)

This Clebsch-Gordan series arises because the vector operators V and W form two = 1 irreducible

subspaces of operators, and when we form T according to Tij = ViWj , we are effectively combining

angular momenta as indicated by Eq. (61). The only difference from our usual practice is that we

are forming products of vector spaces of operators, instead of tensor products of ket spaces.

We have examined this decomposition in the special case Tij = ViWj , but the decomposition

itself applies to any second rank tensor Tij. More generally, Cartesian tensors of any rank 2 arereducible.

It is possible that a given tensorTij may have one or more of the three irreducible components

that vanish. The quadrupole moment tensor (23), for example, is already symmetric and traceless,


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This follows easily from the fact thatKcommutes with any rotation operatorU, and from the fact

that the j = 0 rotation matrices are simply given by the 1 1 matrix (1) (see Eq. (13.61)).Irreducible tensor operators of order 1 are constructed from vector operators by transforming

from the Cartesian basis to the spherical basis. If we let V be a vector operator as defined by

Eq. (13), and define its spherical components by

Vq = T1q =eq V, (64)

then we have

U(R)VqU(R) =eq (R1V) = (Req) V=q

VqD1qq(R), (65)

where we use Eq. (53).

The electric quadrupole operator is given as a Cartesian tensor in Eq. (23). This Cartesian

tensor is symmetric and traceless, so it contains only 5 independent components, which span an

irreducible subspace of operators. In fact, this subspace is associated with angular momentum value

k= 2. It is possible to introduce a set of operators T2q,q=2, . . . , +2 that form a standard angularmomentum basis in this space, that is, that form an order 2 irreducible tensor operator. These can

be regarded as the spherical components of the quadrupole moment tensor. We will explore thissubject in more detail later.

12. Commutation Relations of an Irreducible Tensor Operator with J

Above we presented two equivalent definitions of scalar and vector operators, one involving

transformation properties under rotations, and the other involving commutation relations with J. We

will now do the same with irreducible tensor operators. To this end, we substitute the infinitesimal

form (15) of the rotation operator Uinto both sides of the definition (63).

On the right we will need the D-matrix for an infinitesimal rotation. Since the D-matrix

contains just the matrix elements ofUwith respect to a standard angular momentum basis (this isthe definition of the D-matrices, see Eq. (13.56)), we require these matrix elements in the case of an

infinitesimal rotation. For 1, Eq. (13.56) becomes

Djmm(n, ) =jm|

1 ih

n J|jm= mm i

hjm|n J|jm. (66)

Changing notation (jmm) (kqq) and substituting this and Eq. (15) into the definition (63) ofan irreducible tensor operator, we obtain

1 i

hn J

Tkq

1 +

i

hn J

=q

Tkq

qq ih

kq|n J|kq

, (67)


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or, sincen arbitrary unit vector,

[J, Tkq] =q

Tkqkq|J|kq. (68)

This last equation specifies a complete set of commutation relations of the components ofJ

with the components of an irreducible tensor operator, but it is usually transformed into a different

form. First we take the z -component of both sides and use Jz|kq= hq|kq, so that

kq|Jz |kq= qh qq. (69)

This is Eq. (13.38a) with a change of notation. Then Eq. (68) becomes Eq. (76a) below. Next dotboth sides of Eq. (68) with x iy, and use

J|kq=

(k q)(k q+ 1)h |k, q 1, (70)

or

kq|J|kq=

(k q)(k q+ 1)h q,q1. (71)This is Eq. (13.37b) with a change of notation. Then we obtain Eq. (76b) below. Finally, take the

i-th component of Eq. (68),

[Ji, Tkq] =

q

Tkqkq|Ji|kq, (72)

and form the commutator of both sides with Ji,

[Ji, [Ji, Tkq]] =

q

[Ji, Tkq ]kq|Ji|kq=

qq

Tkqkq|Ji|kqkq|Ji|kq

=q

Tkqkq|J2i |kq,(73)

where we have used Eq. (68) again to create a double sum. Finally summing both sides overi, we

obtain, i

[Ji, [Ji, Tkq]] =

q

Tkqkq|J2|kq. (74)

But kq|J2|kq= k(k+ 1)h2 qq, (75)a version of Eq. (13.38b), so we obtain Eq. (76c) below.

In summary, an irreducible tensor operator satisfies the following commutation relations with

the components of angular momentum:

[Jz , Tkq] = hq T

kq, (76a)

[J, Tkq] = h

(k q)(k q+ 1) Tkq1, (76b)

i

[Ji, [Ji, Tkq]] = h

2k(k+ 1) Tkq. (76c)


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We see that forming the commutator withJ plays the role of a raising or lowering operator for the

components of an irreducible tensor operator. As we did with scalar and vector operators, we can

show that these angular momentum commutation relations are equivalent to the definition (63) of

an irreducible tensor operator. This is done by showing that Eqs. (76) are equivalent to Eq. (63) in

the case of infinitesimal rotations, and that if Eq. (63) is true for any two rotations, it is also true

for their product. Thus by building up finite rotations as products of infinitesimal ones we show

the equivalence of Eqs. (63) and (76). Many books take Eqs. (76) as the definition of an irreducible

tensor operator.

13. The Wigner-Eckart Theorem for Scalar Operators

We now preview the Wigner-Eckart theorem by presenting a special case that is important in its

own right, namely, the case of scalar operators. Recall that the Hamiltonian for any isolated system

is a scalar operator. A typical problem is to find the eigenvalues and eigenstates of the Hamiltonian,

something that usually means working in some basis. In view of the rotational invariance of the

system, a it is logical to use is the standard angular momentum basis. The Wigner-Eckart theorem

for scalars concerns the matrix elements of the scalar operator in this basis.

Let K be the scalar operator, and consider the state K|j m. According to Eq. (10), Kcommutes withJ and hence with any function ofJ, including J2 andJ. Thus we find

J

2

K|j m = h2

j(j+ 1) K|j m,JzK|j m = hm K|j m,

JK|j m = h

(j m)(j m + 1) K|j,m 1. (77)

The first two of these equations show that K|j m is a simultaneous eigenstate of J2 and Jzcorresponding to quantum numbers j and m. But this implies that K|j m must be a linearcombination of the states|j m for the same values ofj and m but possibly different values of.That is, we must be able to write,

K|j m =

|jmCjm, (78)

where the expansion coefficients are Cjm, and where the superscripts and subscripts simply indicate

all the parameters upon which the expansion coefficients could depend. But actually, it turns out

that the expansion coefficients do not depend on m. To show this, we apply raising or lowering

operators to Eq. (78), finding,

JK|j m =K J|j m = h

(j m)(j m+ 1) K|j,m 1=

J|jmCjm

= h

(j m)(j m+ 1)

|j, m 1Cjm, (79)


18/30


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number used in| in Eq. (85). Thus there is a degeneracy, since m can take on the 2j+ 1 valuesm=j, . . . , +j.

15. Energy Levels and Degeneracies in Isolated Systems

Let us apply the results of the previous section to the Hamiltonian of an isolated system. We

replaceKbyH,jn by Enj, and|, the eigenstate ofK, by|njm, showing the quantum numbersupon which it depends. Then we have

|njm=

|j m ujn, (87)

and

H|njm =Enj |njm. (88)Here the index n just labels the eigenvalues of the matrix Cj, which are the energy eigenvalues.

To be definite we can let n label these in ascending order. In many systems the energy spectrum

becomes continuous above some energy, whereupon n must be replaced by a continuous index. The

energy eigenstates are still labeled byj and m as shown, however.

The transformation (87) takes us from an arbitrary standard angular momentum basis to a

standard angular momentum basis that is also an eigenbasis of the Hamiltonian. The basis|njmisa simultaneous eigenbasis of (H, J2, Jz), which must exist because (under our assumption that the

system is isolated) the three operators commute with one another. In addition, the basis vectors

|njm are also linked by raising and lowering operators, connecting states of the same n andj butdifferentm, as explained in Notes 13.

We see that the energy levels of isolated systems come in multiplets of degeneracy 2j+ 1, since

the energy eigenvalues Enj are independent ofm. The physical reason for this is that Jz is the

projection of the angular momentum vector onto the z -axis, so its value depends on the orientation

of the system (or of the axes). Therefore, the quantum number m depends on the orientation. But

the energy does not depend on the orientation, and so must be independent ofm.

We have seen a special case of this already in central force motion, in which the energy eigen-

values are En, where takes the place of the general notation j used here. The energy levels in

central force motion are (2 + 1)-fold degenerate since they do not depend on the magnetic quantum

number. See Sec. 16.4. In central force motion, we have a different radial Schrodinger equation for

each value of. The present discussion concerns any Hamiltonian that is rotationally invariant, and

is more general. Here we find a different matrix Cjfor each value ofj that we must diagonalize.

This is actually equivalent to solving the radial Schrodinger equation for a given, if we identify the

index , which we have been writing as if it were a discrete index, with the continuous index r0 of

the basis of radial wave functions(r r0).This is a far reaching generalization of the results of Sec. 16.4 on the energy levels and degen-

eracies in central force problems, because it applies to any rotationally invariant system of arbitrary


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complexity. It applies, for example, to the aluminum nucleus 27Al, with 13 protons and 14 neutrons,

interacting by means of a Hamiltonian which is not known to high accuracy, but which is known

to be invariant under rotations. That is necessary because the energy of an isolated system cannot

depend on the orientation of the system. In addition, spin and relativistic effects are important in

the dynamics of the nucleons inside the aluminum nucleus. None of this affects the basic conclusion,

that the energy levels are characterized by an angular momentum quantum number j (the quan-

tum number of the total angular momentum of the nucleus), and that the levels are (2j+ 1)-fold

degenerate (because the energies do not depend on m, the quantum number of total Jz).

It is possible that two or more of the eigenvalues of the matrix Cjfor a givenj could coincide,

or that the eigenvalues of one matrix for one value ofj could coincide with those of another value ofj. But in the absence of some systematic symmetry that forces this to happen, it is very unlikely. It

is like choosing two random numbers on some interval and finding that they are exactly equal. Thus,

in typical systems, the energy levels are labeled by a single j value and are (2j + 1)-fold degenerate,

and no more.

In a future version of these notes I will show a plot of the energy levels of some complex nucleus,

to show how they are labeled by an angular momentum quantum number (what we call the spin

of the nucleus, and what we usually write as s instead ofj ). Each of these levels is 2j + 1 (or, in the

usual notation, 2s + 1) fold degenerate. The levels are also characterized by a definite parity (odd or

even), a topic we will take up in Notes 19. The same quantum numbers (spin and parity) also apply

to particles such as the mesons, and for the same reasons (they are isolated systems of quarks).The most important system where some systematic symmetry forces degeneracy among different

j (actually, different ) values is hydrogen, in the electrostatic model. Another example is the

isotropic harmonic oscillator. This point was discussed in Notes 16.

To return to the example of the aluminum nucleus, this is a many-body system with many

bound states, each of which is characterized by an angular momentum quantum number. The

ground state is lowest in energy, and has the angular momentum or spin s = 5/2. When we

studied the behavior of nuclear spins in Notes 14, we assumed the Hilbert space was the (2s+ 1)-

dimensional space spanned by{|sm, m =s , . . . , +s}, for fixed s. Actually this Hilbert space isonly the ground energy eigenspace of the nucleus, regarded as a multiparticle system. We were able

to restrict consideration to that one energy eigenspace and ignore the others, since the magneticenergies we considered in Notes 14 are very small in comparision to the differences in energy between

the ground state and the excited states of the nucleus.

16. Statement and Applications of the Wigner-Eckart Theorem

The Wigner-Eckart theorem is not difficult to remember and it is quite easy to use. In this

section we discuss the statement of the theorem and ways of thinking about it and its applications,

before turning to its proof.

The Wigner-Eckart theorem concerns matrix elements of an irreducible tensor operator with


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respect to a standard angular momentum basis of kets, something we will write in a general notation

asjm|Tkq |j m. As an example of such a matrix element, you may think of the dipole matrixelementsnm|xq|nm that we examined in Sec. 8. In that case the operator (the position ordipole operator) is an irreducible tensor operator with k = 1.

The matrix elementjm|Tkq |j m depends on 8 indices, (jm; j m; kq), and in additionit depends on the specific operator T in question. The Wigner-Eckart theorem concerns the de-

pendence of this matrix element on the three magnetic quantum numbers (mmq), and states that

that dependence is captured by a Clebsch-Gordan coefficient. More specifically, the Wigner-Eckart

theorem states thatjm|Tkq |j mis proportional to the Clebsch-Gordan coefficientjm|jkmq,with a proportionality factor that is independent of the magnetic quantum numbers. That propor-tionality factor depends in general on everything else besides the magnetic quantum numbers, that

is, (j; j ; k) and the operator in question. The standard notation for the proportionality factor is

j||Tk||j, something that looks like the original matrix element except the magnetic quantumnumbers are omitted and a double bar is used. The quantityj||Tk||j is called the reducedmatrix element. With this notation, the Wigner-Eckart theorem states

jm|Tkq |j m=j||Tk||j jm|jkmq.(89)

The reduced matrix element can be thought of as depending on the irreducible tensor operator Tk

and the two irreducible subspaces (

j

) and (j ) that it links. Some authors (for example, Sakurai)include a factor of 1/

2j + 1 on the right hand side of Eq. (89), but here that factor has been

absorbed into the definition of the reduced matrix element. The version (89) is easier to remember

and closer to the basic idea of the theorem.

To remember the Clebsch-Gordan coefficient it helps to suppress the bra jm| from thematrix element and think of the ket Tkq |j m, or, more precisely, the (2j + 1)(2k+ 1) kets thatare produced by letting m and qvary over their respective ranges. This gives an example of an

operator with certain angular momentum indices multiplying a ket with certain angular momentum

indices. It turns out that such a product of an operator times a ket has much in common with the

product (i.e., the tensor product) of two kets, insofar as the transformation properties of the product

under rotations are concerned. That is, suppose we were multiplying a ket|kq with the givenangular momentum quantum numbers times another ket|jm with different angular momentumquantum numbers. Then we could find the eigenstates of total angular momentum by combining

the constituent angular momenta according to kj . Actually, in thinking of kets Tkq |jm, it iscustomary to think of the product of the angular momenta in the reverse order, that is, j k. Thisis an irritating convention because it makes the Wigner-Eckart theorem harder to remember, but I

suspect it is done this way because in practicek tends to be small and j large.

In any case, thinking of the product of kets, the product

|jm |kq=|jkmq (90)


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contains various components of total J2 andJz , that is, it can be expanded as a linear combination

of eigenstates of total J2 and Jz, with expansion coefficients that are the Clebsch-Gordan coeffi-

cients. The coefficient with total angular momentumj andz -componentm is the Clebsch-Gordan

coefficientjm|jkmq, precisely what appears in the Wigner-Eckart theorem (89).Probably the most useful application of the Wigner-Eckart theorem is that it allows us to easily

write down selection rules for the given matrix element, based on the selection rules of the Clebsch-

Gordan coefficient occurring in Eq. (89). In general, a selection ruleis a rule that tells us when a

matrix element must vanish on account of some symmetry consideration. The Wigner-Eckart the-

orem provides us with all the selection rules that follow from rotational symmetry; a given matrix

element may have other selection rules based on other symmetries (for example, parity). The selec-tion rules that follow from the Wigner-Eckart theorem are that the matrix elementj m|Tkq |j mvanishes unless m =m+ qand j takes on one of the values,|j k|, |j k| + 1, . . . , j+ k.

Furthermore, suppose we actually have to evaluate the matrix elementsjm|Tkq |j m forall (2k + 1)(2j + 1) possibilities we get by varying q and m. We must do this, for example, in

computing atomic transition rates. (We need not vary m independently, since the selection rules

enforcem =m+q.) Then the Wigner-Eckart theorem tells us that we actually only have to do one of

these matrix elements (presumably, whichever is the easiest), because if we know the left hand side of

Eq. (89) for one set of magnetic quantum numbers, and if we know the Clebsch-Gordan coefficient on

the right-hand side, then we can determine the proportionality factor, that is, the reduced matrix

element. Then all the other matrix elements for other values of the magnetic quantum numbersfollow by computing (or looking up) Clebsch-Gordan coefficients. This procedure requires that the

first matrix element we calculate be nonzero.

In some other cases, we have analytic formulas for the reduced matrix element. That was the

case of the application in Sec. 8, where the three-Ym formula allowed us to compute the propor-

tionality factor explicitly.

In the special case of a scalar operator, T00 =K, we have k = q= 0, and the Clebsch-Gordan

coefficient is

jm|j0m0 =jjmm. (91)Thus the Wigner-Eckart theorem becomes

jm|K|j m=j||K||j jjmm, (92)

which is equivalent to Eq. (83) with the reduced matrix element written as Cj.

17. Proof of the Wigner-Eckart Theorem

Consider the product of kets|jm |kq=|jkmqwith the given angular momentum quantumnumbers, and consider the (2j +1)(2k +1)-dimensional product space spanned by such kets when we

allow the magnetic quantum numbersm and qto vary over their respective ranges. The eigenstates


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|JM of total J2 and Jz in this space are given by the Clebsch-Gordan expansion,

|JM=mq

|jk mqjkmq|JM. (93)

Moreover, the states|JM for fixed J and M =J , . . . , +Jform a standard angular momentumbasis in an invariant, irreducible subspace of dimension 2J+1 in the product space. This means that

the basis states|JMare not only eigenstates of total J2 andJz, but they are also linked by raisingand lowering operators. Equivalently, the states|JM transform as a standard angular momentumbasis under rotations,

U|JM= M

|JM

DJMM(U). (94)

Now consider the (2j+ 1)(2k+ 1) kets Tkq |j m obtained by varying m and q. We constructlinear combinations of these with the same Clebsch-Gordan coefficients as in Eq. (93),

|X; JM=mq

Tkq |j mjkmq|JM, (95)

and define the result to be the ket |X; JM, as indicated. The indices JM in the ket|X; JMindicate that the left-hand side depends on these indices, because the right hand side does; initially

we assume nothing else about this notation. Similarly, X simply stands for everything else the

left-hand side depends on, that is, Xis an abbreviation for the indices (kj).However, in view of the similarity between Eqs. (93) and (95), we can guess that|X; JM is

actually an eigenstate ofJ2 and Jz with quantum numbers J and M, and that the states|X; JMare related by raising and lowering operators. That is, we guess

Jz|X; JM =Mh |X; JM, (96a)J|X; JM =

(J M)(J M+ 1)h |X; J, M 1, (96b)

J2|X; JM =J(J+ 1)h2 |X; JM. (96c)

If true, this is equivalent to the transformation law,

U|X; JM = M

|X; JM DJMM(U), (97)

exactly as in Eq. (94). Equations (96) and (97) are equivalent because Eq. (96) can be obtained

from Eq. (97) by specializing to infinitesimal rotations, while Eq. (97) can be obtained from Eq. (96)

by building up finite rotations out of infinitesimal ones.

In Sec. 18 below we will prove that these guesses are correct. For now we merely explore the

consequences. The logic is a generalization of what we applied earlier in the case of scalar operators

in Sec. 13. To begin, since|X; JMis an eigenstate ofJ2 andJz with quantum numbers JandM,it can be expanded as a linear combination of the standard basis kets|j m with the same values


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j = J and m = M, but in general all possible values of. That is, we have an expansion of the

form,

|X; JM=

|JM CkJMj , (98)

where the indices on the expansion coefficients CkJMj simply list all the parameters they can depend

on. These coefficients, do not, however, depend on M, as we show by applying raising or lowering

operators to both sides, and using Eq. (96b). This gives

(J M)(J M+ 1)h |X; J, M 1

=

(J M)(J M+ 1)h |J, M 1 CkJMj , (99)

or, after cancelling the square roots,

|X; J, M 1=

|J, M 1 CkJMj . (100)

Comparing this to Eq. (98), we see that the expansion coefficients are the same for all M values,

and thus independent ofM. We will henceforth write simply CkJj for them.

Now we return to the definition (95) of the kets |X; JM and use the orthogonality of theClebsch-Gordan coefficients (17.33) to solve for the kets Tkq

|j m

. This gives

Tkq |j m=JM

|X; JMJM|jk mq=JM

|JM CkJjJM|jkmq, (101)

where we use Eq. (98), replacing with . Now multiplying this byjm| and using theorthonormality of the basis|j m, we obtain

jm|Tkq |j m= Ckjj

jm|jkmq, (102)

which is the Wigner-Eckart theorem (89) if we identify

Ckjj

=

j

||Tk

||j

. (103)

18. Proof of Eq. (97)

To complete the proof of the Wigner-Eckart theorem we must prove Eq. (97), that is, we

must show that the kets|X; JM transform under rotations like the vectors of a standard angularmomentum basis. To do this we call on the definition of|X; JM, Eq. (95), and apply U to bothsides,

U|X; JM =mq

U TkqU U|j mjkmq|JM. (104)


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Next we use the definition of an irreducible tensor operator (63) and the transformation law for

standard basis vectors under rotations, Eq. (13.79), to obtain

U|X; JM =mq

mq

Tkq |j m Djmm(U) Dkqq(U) jkmq|JM. (105)

We now call on Eq. (17.40) with a change of indices,

Djmm(U)Dkqq(U) =

JMM

jkmq|JM DJMM(U) JM|jk mq, (106)

which expresses the product ofD-matrices in Eq. (105) in terms of single D-matrices. When we

substitute Eq. (106) into Eq. (105), the mq-sum is doable by the definition (95),

mq

Tkq |j mjkmq|JM=|X; JM, (107)

and the mq-sum is doable by the orthogonality of the Clebsch-Gordan coefficients,

mq

JM|jkmqjkmq|JM= JJMM. (108)

Altogether, Eq. (105) becomes

U|X; JM = JMM

|X; JMDJMM(U) JJMM = M

|X; JMDJMM(U). (109)

This proves Eq. (97).

Instead of proving Eq. (97), many authors (for example, Sakurai) prove the equivalent set

of statements (96), which involve the actions of the angular momentum operators on the states

|X; JM. I think the transformation properties under rotations are little easier. In either case, therest of the proof is the same.

19. Products of Irreducible Tensor Operators

As we have seen, the idea behind the Wigner-Eckart theorem is that a product of an irreducibletensor operatorTkq times a ket of the standard basis|j mtransforms under rotations exactly as thetensor product of two kets of standard bases with the same quantum numbers,|jm |kq. Similarly,it turns out that the product of two irreducible tensor operators, say, Xk1q1Y

k2q2

, transforms under

rotations exactly like the tensor product of kets with the same quantum numbers, |k1q1 |k2q2.In particular, such a product of operators can be represented as a linear combination of irreducible

tensor operators with order k lying in the range|k1k2|, . . . , k1 + k2, with coefficients that areClebsch-Gordan coefficients. That is, we can write

Xk1q1 Yk2q2

=kq

Tkqkq|k1k2q1q2, (110)


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where the Tkq are new irreducible tensor operators.

To prove this, we first solve for Tkq,

Tkq =q1q2

Xk1q1Yk2q2

k1k2q1q2|kq, (111)

which we must show is an irreducible tensor operator. To do this, we conjugate both sides of this

with a rotation operator Uand use the fact that Xand Y are irreducible tensor operators,

U TkqU =

q1q2

U Xk1q1U U Yk2q2 U

k1k2q1q2|kq

=q1q2

q1q2

Xk1q1

Yk2q2

Dk1q1q1

(U)Dk2q2q2

(U) k1k2q1q2|kq. (112)

Next we use Eq. (17.40) with a change of symbols,

Dk1q1q1

(U)Dk2q2q2

(U) =KQQ

k1k2q1q2|KQDKQQ(U)KQ|k1k2q1q2, (113)

which we substitute into Eq. (112). Then theq1q2-sum is doable in terms of the expression (111)

forTkq,

q1q2

Xk1q1

Yk2q2

k1k2q1q2

|KQ

= TKQ, (114)

and the q1q2-sum is doable by the orthogonality of the Clebsch-Gordan coefficients,

q1q2

KQ|k1k2q1q2k1k2q1q2|kq= KkQq. (115)

Then Eq. (112) becomes

U TkqU =

KQQ

TKQDKQQ KkQq =

q

TkqDkqq(U). (116)

This shows thatTkq is an irreducible tensor operator, as claimed.

As an application, two vector operators V and W, may be converted into k = 1 irreducibletensor operators Vq and Wq by going over to the spherical basis. From these we can construct

k= 0, 1, 2 irreducible tensor operators according to

Tkq =q1q2

Vq1Wq211q1q2|kq. (117)

This will yield the same decomposition of a second rank tensor discussed in Sec. 10, where we found

a scalar (k= 0), a vector (k= 1), and a symmetric, traceless tensor (k= 2).


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Problems

1. The following problem will help you understand irreducible tensor operators better. LetE bea ket space for some system of interest, and let A be the space of linear operators that act onE.For example, the ordinary Hamiltonian is contained inA, as are the components of the angularmomentum J, the rotation operators U(R), etc. The spaceA is a vector space in its own right,

just likeE; operators can be added, multiplied by complex scalars, etc. Furthermore, we may beinterested in certain subspaces ofA, such as the 3-dimensional space of operators spanned by thecomponentsVx, Vy, Vz of a vector operatorV.

Now letSbe the space of linear operators that act onA. We call an element ofSa superoperator because it acts on ordinary operators; ordinary operators inA act on kets inE. We willdenote super-operators with a hat, to distinguish them from ordinary operators. (This terminology

has nothing to do with supersymmetry.)

Given an ordinary operatorA A, it is possible to associate it in several different ways with asuper-operator. For example, we can define a super operator AL, which acts by left multiplication:

ALX= AX, (118)

whereX is an arbitrary ordinary operator. This equation obviously defines a linear super-operator,

that is, AL(X+ Y) = ALX+ ALY, etc. Similarly, we can define a super-operator associated with

A by means of right multiplication, or by means of the forming of the commutator, as follows:

ARX= X A,

ACX= [A, X].(119)

There are still other ways of associating an ordinary operator with a super-operator. Let R be a

classical rotation, and letU(R) be a representation of the rotations acting on the ket spaceE. Thus,the operatorsU(R) belong to the spaceA. Now associate such a rotation operatorU(R) inAwitha super-operatorU(R) inS, defined by

U(R)X= U(R) X U(R). (120)

Again, U(R) is obviously a linear super-operator.

(a) Show that U(R) forms a representation of the rotations, that is, that

U(R1)U(R2) = U(R1R2). (121)

This is easy.

Now let U(R) be infinitesimal as in Eq. (15), and let

U(R) = 1 ih

n J. (122)


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(Here the hat on n denotes a unit vector, while that on J denotes a super-operator.) Express the

super-operator J in terms of ordinary operators. Write Eqs. (76) in super-operator notation. Work

out the commutation relations of the super-operatorsJ.

(b) Now write out nine equations, specifying the action of the three super-operators Ji on the the

basis operators Vj . Write the answers as linear combinations of the Vj s. Then write out six more

equations, specifying the action of the super raising and lowering operators, J, on the three Vj.

Now find the operatorAthat is annihilated by J+. Do this by writing out the unknown operator

as a linear combination of the Vj s, in the form

A= axVx+ ayVy+ azVz , (123)

and then solving for the coefficients ai. Show that this operator is an eigenoperator ofJz with

eigenvalue +h. In view of these facts, the operatorA must be a stretched operator for k = 1;

henceforth write T11 for it. This operator will have an arbitrary, complex multiplicative constant,

call itc. Now apply J, and generateT10 andT11. Choose the constantc to makeT

10 look as simple

as possible. Then write

T1q =eq V, (124)and thereby discover the spherical basis.

2. This problem concerns quadrupole moments and spins. It provides some background for the

following problem.

(a) In the case of a nucleus, the spin Hilbert spaceEspin = span{|sm, m=s , . . . , +s} is actuallythe ground state of the nucleus. It is customary to denote the angular momentum j of the ground

state bys. This state is (2s +1)-fold degenerate. The nuclear spin operatorS is really the restriction

of the total angular momentum of the nucleus Jto this subspace of the (much larger) nuclear Hilbert

space.

LetAkq andBkq be two irreducible tensor operators onEspin. As explained in these notes, when

we say irreducible tensor operator we are really talking about the collection of 2k+ 1 operators

obtained by setting q=k , . . . , +k. Use the Wigner-Eckart theorem to explain why any two such

operators of the same orderk are proportional to one another. This need not be a long answer.Thus, all scalars are proportional to a standard scalar (1 is convenient), and all vector operators

(for example, the magnetic moment ) are proportional to a standard vector (S is convenient), etc.

For a givens, what is the maximum value ofk? What is the maximum order of an irreducible

tensor operator that can exist on space Espinfor a proton (nucleus of ordinary hydrogen)? A deuteron(heavy hydrogen)? An alpha particle (nucleus of helium)?

(b)Let AandBbe two vector operators (on any Hilbert space, not necessarily Espin), with sphericalcomponents Aq, Bq, as in Eq. (64). As explained in the notes, Aq and Bq are k = 1 irreducible

tensor operators. As explained in Sec. 19, it is possible to construct irreducible tensor operators


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Tkq for k = 0, 1, 2 out of the nine operators,{AqBq , q , q =1, 0, 1}. Write out the nine operatorsT00 , T

11 , T

10 , T

11, T

22 , . . . , T

22 in terms of the Cartesian products AiBj. Just look up the Clebsch-

Gordan coefficients. Show thatT00 is proportional to A B, thatT1q is proportional to the sphericalcomponents ofAB, and that T2q specifies the components of the symmetric and traceless part ofthe Cartesian tensor AiBj , which is

1

2(AiBj+ AjBi) 1

3(A B)ij . (125)

(c ) In classical electrostatics, the quadrupole moment tensor Qij of a charge distribution (x) is

defined byQij =

d3x (x)[3xixj r2 ij], (126)

where x is measured relative to some origin inside the charge distribution. The quadrupole mo-

ment tensor is a symmetric, traceless tensor. The quadrupole energy of interaction of the charge

distribution with an external electric field E =is

Equad=1

6

ij

Qij2(0)

xixj. (127)

This energy must be added to the monopole and dipole energies, plus the higher multipole energies.

In the case of a nucleus, we choose the origin to be the center of mass, whereupon the dipole

moment and energy vanish. The monopole energy is just the usual Coulomb energyq(0), whereq

is the total charge of the nucleus. Thus, the quadrupole term is the first nonvanishing correction.

However, the energy must be understood in the quantum mechanical sense.

Let{x, = 1, . . . , Z }be the position operators for the protons in a nucleus. The neutrons areneutral, and do not contribute to the electrostatic energy. The electric quadrupole moment operator

for the nucleus is defined by

Qij =e

(3xixj r2 ij), (128)

where e is the charge of a single proton. In an external electric field, the nuclear Hamiltonian

contains a term Hquad, exactly in the form of Eq. (127), but now interpreted as an operator.

The operator Qij, being symmetric and traceless, constitutes the Cartesian specification of ak= 2 irreducible tensor operator, that you could turn into standard form T2q , q=2, . . . , +2 usingthe method of part (b) if you wanted to. Well say with the Cartesian form here, however. When

the operatorQij is restricted to the ground state (really a manifold of 2s + 1 degenerate states), it

remains ak = 2 irreducible tensor operator. According to part (a), it must be proportional to some

standard irreducible tensor operator, for which 3SiSj S2ij is convenient. That is, we must beable to write

Qij =a(3SiSj S2ij), (129)for some constant a.


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18 - Irreducible Tensor Operators and the Wigner-Eckart Theorem.pdf